Lecture 16 Strongly Connected Components CS473-Algorithms I Lecture 16 Strongly Connected Components CS 473 Lecture 16

Strongly Connected Components Definition: a strongly connected component (SCC) of a directed graph G(V,E) is a maximal set of vertices U V such that For each u,v U we have both u  v and v  u i.e., u and v are mutually reachable from each other (u  v) Let GT(V,ET) be the transpose of G(V,E) where ET {(u,v): (u,v)  E} i.e., ET consists of edges of G with their directions reversed Constructing GT from G takes O(V+E) time (adjacency list rep) Note: G and GT have the same SCCs (u  v in G u  v in GT) CS 473 Lecture 16

Strongly Connected Components Algorithm Run DFS(G) to compute finishing times for all uV Compute GT Call DFS(GT) processing vertices in main loop in decreasing f[u] computed in Step (1) Output vertices of each DFT in DFF of Step (3) as a separate SCC CS 473 Lecture 16

Strongly Connected Components Lemma 1: no path between a pair of vertices in the same SCC, ever leaves the SCC Proof: let u and v be in the same SCC  u  v let w be on some path u w  v  u w but v  u   a path w v  u  w u therefore u and w are in the same SCC u SCC w v QED CS 473 Lecture 16

SCC: Example CS 473 Lecture 16

SCC: Example Run DFS(G) to compute finishing times for all uV CS 473 Lecture 16

SCC: Example Run DFS(G) to compute finishing times for all uV CS 473 Lecture 16

SCC: Example Run DFS(G) to compute finishing times for all uV CS 473 Lecture 16

SCC: Example b, e, a, c, d, g, h, f  Vertices sorted according to the finishing times: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example Compute GT CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (3) Call DFS(GT) processing vertices in main loop in decreasing f[u] order: b, e, a, c, d, g, h, f  CS 473 Lecture 16

SCC: Example (4) Output vertices of each DFT in DFF as a separate SCC Cc{c,d} Cb{b,a,e} Cg{g,f} Ch{h} CS 473 Lecture 16

SCC: Example Cc Cb Cg Ch CS 473 Lecture 16

Strongly Connected Components Thrm 1: in any DFS, all vertices in the same SCC are placed in the same DFT Proof: let r be the first vertex discovered in SCC Sr because r is first, color[x]WHITE xSr{r} at time d[r] So all vertices are WHITE on each r  x path xSr{r} since these paths never leave Sr Hence each vertex in Sr{r}becomes a descendent of r (White-path Thrm) G at time d[r] r W W W Sr QED CS 473 Lecture 16

Notation for the Rest of This Lecture d[u] and f[u] refer to those values computed by DFS(G) at step (1) u  v refers to G not GT Definition: forefather  (u) of vertex u  (u)  That vertex w such that u  w and f[u] is maximized  (u)  u possible because u  u  f[u]  f[ (u)] CS 473 Lecture 16

Strongly Connected Components Lemma 2:  ( (u))   (u) Proof try to show that f[ ( (u))]  f[ (u)] : For any u,v V; u  v  Rv  Ru  f[ (v)]  f[ (u)] So, u   (u)  f[ ( (u))]  f[ (u)] Due to definition of  (u) we have f[ ( (u))]  f[ (u)] Therefore f[ ( (u))]  f[ (u)] QED Note: f[x]  f[y]  x  y (same vertex) Rv {w: v w} u v  v  (u) CS 473 Lecture 16

Strongly Connected Components Properties of forefather: Every vertex in an SCC has the same forefather which is in the SCC Forefather of an SCC is the representative vertex of the SCC In the DFS of G, forefather of an SCC is the first vertex discovered in the SCC last vertex finished in the SCC CS 473 Lecture 16

Strongly Connected Components THM2: (u) of any u  V in any DFS of G is an ancestor of u PROOF: Trivial if (u)  u. If (u)  u, consider color of (u) at time d[u] (u) is GRAY: (u) is an ancestor of u  proving the theorem (u) is BLACK: f [(u)] < f [u]  contradiction to def. of (u) (u) is WHITE:  2 cases according to colors of intermediate vertices on p(u, (u)) Path p(u, (u)) at time d[u]: CS 473 Lecture 16

Strongly Connected Components Case 1: every intermediate vertex xi  p(u, (u)) is WHITE (u) becomes a descendant of u (WP-THM)  f [(u)] < f [u]  contradiction Case 2:  some non-WHITE intermediate vertices on p(u, (u)) Let xt be the last non-WHITE vertex on p(u, (u))  u, x1, x2,…, xr, (u) Then, xt must be GRAY since BLACK-to-WHITE edge (xt, xt+1) cannot exist But then, p(xt, (u))   xt+1, xt+2,…, xr, (u) is a white path  (u) is a descendant of xt (by white-path theorem) f [xt] > f [(u)] contradicting our choice for (u) Q.E.D. CS 473 Lecture 16

Strongly Connected Components C1: in any DFS of G  (V, E) vertices u and (u) lie in the same SCC, u  V PROOF: u  (u) (by definition) and (u)  u since (u) is an ancestor of u (by THM2) THM3: two vertices u,v V lie in the same SCC  (u) = (v) in a DFS of G  (V, E) PROOF: let u and v be in the same SCC Cuv  u  v CS 473 Lecture 16

Strongly Connected Components w: v  w  u  w and w: u  w  v  w, i.e., every vertex reachable from u is reachable from v and vice-versa So, w  (u)  w  (v) and w  (v)  w  (u) by definition of forefather PROOF: Let (u)  (v)  wCw  uCw by C1 and vCw by C1 By THM3: SCCs are sets of vertices with the same forefather By THM2 and parenthesis THM: A forefather is the first vertex discovered and the last vertex finished in its SCC CS 473 Lecture 16

SCC: Why do we Run DFS on GT? Consider r  V with largest finishing time computed by DFS on G r must be a forefather by definition since r  r and f [r] is maximum in V Cr  ?: Cr  vertices in r’s SCC  {u in V: (u)  r}  Cr  {u  V: u  r and f [x]  f [r] x  Ru} where Ru {v  V: u  v}  Cr  {u  V: u  r} since f [r] is maximum  Cr  RrT  {u  V: r  u in GT}  reachability set of r in GT i.e., Cr  those vertices reachable from r in GT Thus DFS-VISIT(GT, r) identifies all vertices in Cr and blackens them CS 473 Lecture 16

SCC: Why do we Run DFS on GT? BFS(GT, r) can also be used to identify Cr Then, DFS on GT continues with DFS-VISIT(GT, r ) where f [r ] > f [w] w  V Cr r must be a forefather by definition since r  r and f [r ] is maximum in V Cr CS 473 Lecture 16

SCC: Why do we Run DFS on GT? Hence by similar reasoning DFS-VISIT(GT, r ) identifies Cr Impossible since otherwise r, w  Cr  r, w would have been blackened Thus, each DFS-VISIT(GT, x) in DFS(GT) identifies an SCC Cx with   x CS 473 Lecture 16